 Solution

The occurrence of $y$ in $P(x,y)$ lies under the quantifier $\exists x$. Directly inserting the term $x+z$ would capture its free $x$: the result $\exists xP(x,x+z)$ would make the inserted $x$ bound and change the intended dependence on the original free value. Therefore the term is not free for the literal substitution requested.

Perform capture-avoiding first-order substitution by first renaming bound variables to fresh names. Renaming the first binder to $u$ is necessary; renaming the second to $v$ is optional but makes the result especially clear. These renamings give the equivalent formula $(\exists uP(u,y))\vee (\exists vQ(v))$. Substituting only the free occurrences of $y$ now givesBoth $x$ and $z$ in the inserted term remain free variables. The original $\exists z$ has scope only over its own disjunct and would not capture the inserted $z$ in the first disjunct; its optional renaming is not a necessary capture repair. This distinction avoids confusing reuse of a letter elsewhere with binding within the relevant scope.